Why Generate All Combinations Sum K in DSA Java?

The Big Idea

What if you could instantly find every possible group that adds up to your target without missing any?

The Scenario

Imagine you want to find all groups of friends whose ages add up to exactly 30 years. You try to list every possible group by hand, checking each combination one by one.

The Problem

Doing this manually is slow and confusing because the number of possible groups grows very fast. You might miss some groups or repeat others, making it error-prone and frustrating.

The Solution

Using a method to generate all combinations that sum to a target number helps you find every possible group quickly and without mistakes. It automatically explores all options and keeps track of sums efficiently.

Before vs After

✗ Before

int[] ages = {5, 10, 15, 20};
// Manually check each group if sum == 30
// Too many checks and easy to miss combinations

✓ After

void findCombinations(int[] ages, int target, List<Integer> current, int start) {
  if (target == 0) {
    System.out.println(current);
    return;
  }
  for (int i = start; i < ages.length; i++) {
    if (ages[i] <= target) {
      current.add(ages[i]);
      findCombinations(ages, target - ages[i], current, i + 1);
      current.remove(current.size() - 1);
    }
  }
}

What It Enables

This lets you quickly find all groups of numbers that add up to a target, enabling powerful problem solving in many areas like budgeting, planning, and games.

Real Life Example

Planning a party where you want to select dishes with prices adding up exactly to your budget. This method helps find all possible dish combinations that fit your budget.

Key Takeaways

Manual checking of combinations is slow and error-prone.

Generating all combinations with a target sum automates and simplifies the process.

This approach helps solve real-world problems involving selection and budgeting.

Practice

1. What does the problem "Generate All Combinations Sum K" ask you to find?

easy

A. All unique combinations of numbers that add up to K

B. The largest number less than K

C. The sum of all numbers from 1 to K

D. The product of all numbers that divide K

Solution

1. Step 1: Understand the problem goal

   The problem asks to find all unique sets of numbers that add exactly to the target sum K.

2. Step 2: Clarify what combinations mean

   Combinations mean selecting numbers without caring about order, but each combination must sum to K.

3. Final Answer:

   All unique combinations of numbers that add up to K -> Option A

4. Quick Check:

   Sum of combinations = K [OK]

Hint: Focus on sums equal to K, not order or multiplication [OK]

Common Mistakes:

• Confusing combinations with permutations
• Looking for product instead of sum
• Ignoring uniqueness of combinations

2. Which of the following Java method signatures is correct for a function that generates all combinations summing to K using backtracking?

easy

A. public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start)

B. public int findCombinations(int[] nums, int target)

C. void findCombinations(int target, int[] nums, List result)

D. public List<Integer> findCombinations(int target, int[] nums)

Solution

1. Step 1: Identify parameters needed for backtracking

   We need the array, target sum, current combination list, result list, and start index to avoid duplicates.

2. Step 2: Match method signature with these parameters

   public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) matches all needed parameters and returns void as it fills the result list.

3. Final Answer:

   public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) -> Option A

4. Quick Check:

   Correct parameters for backtracking method [OK]

Hint: Backtracking needs current, result lists, and start index [OK]

Common Mistakes:

• Missing start index parameter
• Wrong return type for backtracking method
• Incorrect parameter order

3. What will be the output of the following Java code snippet?

List result = new ArrayList<>();
int[] nums = {2, 3, 6, 7};
int target = 7;
findCombinations(nums, target, new ArrayList<>(), result, 0);
System.out.println(result);

Assuming findCombinations correctly finds all unique combinations summing to target.

medium

A. [[2, 3, 6], [7]]

B. [[7]]

C. [[2, 2, 3], [7]]

D. [[2, 2, 2, 3], [7]]

Solution

1. Step 1: Identify combinations that sum to 7

   From nums, 7 alone sums to 7, and 2+2+3 also sums to 7.

2. Step 2: Confirm no other combinations sum to 7

   3+6=9, 2+3+6=11, 2+2+2+3=9 (too large), so only [2,2,3] and [7] are valid.

3. Final Answer:

   [[2, 2, 3], [7]] -> Option C

4. Quick Check:

   Valid combinations sum to 7 [OK]

Hint: Check sums carefully; only valid combos appear [OK]

Common Mistakes:

• Including invalid sums
• Missing repeated elements allowed
• Confusing order of elements

4. Identify the error in this Java backtracking code snippet for generating combinations summing to K:

void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) {
    if (target == 0) {
        result.add(new ArrayList<>(current));
        return;
    }
    for (int i = start; i < nums.length; i++) {
        if (nums[i] > target) continue;
        current.add(nums[i]);
        findCombinations(nums, target - nums[i], current, result, i);
        current.remove(current.size());
    }
}

medium

A. The recursive call should use i + 1 instead of i

B. The base case should check if target < 0 instead of target == 0

C. The loop should start from 0 instead of start

D. Removing element with current.remove(current.size()) causes IndexOutOfBoundsException

Solution

1. Step 1: Check the remove method usage

   List indices start at 0, so last element index is current.size() - 1, not current.size().

2. Step 2: Identify correct remove call

   Using current.remove(current.size()) causes IndexOutOfBoundsException because index is out of range.

3. Final Answer:

   Removing element with current.remove(current.size()) causes IndexOutOfBoundsException -> Option D

4. Quick Check:

   List remove index must be size-1 [OK]

Hint: Remove last element with index size-1, not size [OK]

Common Mistakes:

• Removing element at wrong index
• Ignoring base case correctness
• Wrong loop start index

5. Given the array nums = {1, 2, 3} and target sum 4, which of the following is the correct set of all unique combinations where numbers can be repeated and sum to 4?

hard

A. [[1,1,1,1], [1,1,2], [1,3], [2,2], [3,1]]

B. [[1,1,1,1], [1,1,2], [1,3], [2,2]]

C. [[1,1,1,1], [1,1,2], [1,3], [2,2], [4]]

D. [[1,1,1,1], [1,1,2], [2,2], [3,1]]

Solution

1. Step 1: List all combinations summing to 4 with repeats allowed

   Possible combos: [1,1,1,1], [1,1,2], [1,3], [2,2]. Note [3,1] is same as [1,3] and duplicates are not allowed.

2. Step 2: Check options for duplicates and invalid elements

   [[1,1,1,1], [1,1,2], [1,3], [2,2], [3,1]] includes [3,1] duplicate of [1,3]. [[1,1,1,1], [1,1,2], [1,3], [2,2], [4]] includes [4] which is not in nums. [[1,1,1,1], [1,1,2], [2,2], [3,1]] misses [1,3]. [[1,1,1,1], [1,1,2], [1,3], [2,2]] correctly lists unique combinations.

3. Final Answer:

   [[1,1,1,1], [1,1,2], [1,3], [2,2]] -> Option B

4. Quick Check:

   Unique combos sum to 4 without duplicates [OK]

Hint: Avoid duplicates by sorting or fixed order in combos [OK]

Common Mistakes:

• Counting duplicates as separate combinations
• Including numbers not in the array
• Ignoring repeated use of numbers